# Properties of Dedekind Domains

Throughout this article, *A* denotes a Dedekind domain.

Proposition 1.Every fractional ideal of A can be written as

where each is a maximal ideal. The expression is unique up to permutation of terms.

**Note**

In particular, the maximal ideals of *A* generate its Picard group.

**Proof**

**Existence**. First suppose is a non-zero ideal. If we are done. Otherwise it is contained in some maximal ideal , necessarily invertible. Now , because if equality holds multiplying both sides by gives , so . We replace by and repeat the process; this cannot continue indefinitely or we would have

Hence is a finite product of maximal ideals. Now a general fractional ideal is of the form for a non-zero ideal and . Write both and as a product of maximal ideals and we are done.

**Uniqueness**. Suppose , where (some possibly zero). If , after cancelling terms and reordering, we obtain a relation of the form:

This is an equality of ideals. We get a contradiction because the LHS is contained in but the RHS is not. ♦

**Exercise A**

Prove that an integral domain is a PID if and only if it is a Dedekind domain and a UFD. Hence we have the following Venn diagram.

Now we have the following.

Proposition 2.Let M, N be non-zero fractional ideals of A. Write, as in proposition 1,

where each is maximal and . Then

**Proof**

Exercise. [ Localize at each ; reduce to the case where *A* is a dvr. ]

# Constructing Dedekind Domains

How do we find Dedekind domains? For starters, a PID is a Dedekind domain so we have and (for any field *k*), but these are not too interesting.

Here is a standard way to construct new Dedekind domains from an existing one *A*. Let *K* be its field of fractions and *L* be a finite extension of *K*; we let *B* be the integral closure of *A* in *L*.

**Exercise B**

Prove that *L* is the field of fractions of *B*.

Now *B* has the following properties:

- it is a normal domain by construction;
- dim
*B*= 1 since*B*is an integral extension of*A*, and dim*A*= 1 (by main theorem here).

Hence, if we can show that *B* is noetherian, then it is a Dedekind domain. But this result, called the *Krull-Akizuki theorem*, is surprisingly hard to prove. In most cases of interest, *B* will turn out to be a finitely generated *A*-module; thus *B* is noetherian as an *A*-module and hence as a module over itself.

**Examples**

1. Let be a finite extension and be the integral closure of in *K*. One can show that is a discrete subgroup of , when we topologize by identifying it with . Hence is a finitely generated -module; in particular it is a noetherian ring and hence a Dedekind domain. A more general result of this nature will be proven later.

2. Let *k* be any field such that . The integral closure of in is . [ Proof: exercise. ] Since *A* is clearly noetherian, it is a Dedekind domain.

**Exercise C**

Let be a square-free integer and , which is quadratic over . Prove that

# Valuation

Definition.Let . Each maximal gives a function

,

which takes to the exponent of in the above factorization of . This is called the

valuationof at .

**Exercise D**

1. Prove the following properties of : for any we have:

.

Verify that if we set these properties hold for all .

2. Define a distance function on *K *by . Prove that (*K*, *d*) is a metric space, where the metric satisfies the **strong triangular inequality**.

A metric which satisfies this inequality is called an **ultrametric**. Interpret and prove the following statement: *in an ultrametric space, every triangle is isoceles*.

## Valuation in Geometry

Suppose , a Dedekind domain. Recall that points *P* on the curve correspond bijectively to maximal ideals . Then is just the order of vanishing of *f *at *P*.

For example, let where ; for convenience write . Let us compute for :

Now since *f* does not vanish on *P*. From an earlier example, we have so and .

As a simple exercise, compute .

# Composition Series

Recall in the last example here that for distinct maximal ideals of any ring *A* and we have a ring isomorphism

This gives the following.

Proposition 3.If A is a Dedekind domain with finitely many maximal ideals , then it is a PID.

**Proof**

It suffices to prove that each is principal. Let us consider . Since we can pick . From the isomorphism

,

we can find such that

Then *x* is divisible by but not its square, and not divisible by . Thus . ♦

Proposition 4.Let be a non-zero ideal; write

Then the composition factors of as an A-module comprise of copies of for . In particular

.

**Note**

Recall that the composition factors of as an *A*-module are identical to those as a module over itself.

**Proof**

The above ring isomorphism

is in fact an isomorphism of *A*-algebras. Hence . We obtain a series of submodules of

with successive quotients . It suffices to show each of these is of dimension 1 over . To prove that, we recall that is a dvr so is a principal ideal. Since is already a module over we have:

as desired. ♦

**Note**

Consider the special case where is the coordinate ring of a variety *V* over an algebraically closed field *k*. If *A* is a Dedekind domain, then since for each maximal ideal we have

.